Tag Info

New answers tagged

3

The star acts as an "effective point source" since it is so far away and its angular extent is so small - in other words, the optical signal arriving at earth is "very nearly" a plane wave with the same phase over a large extent. This is what enables us to do interferometry. Think of waves, not photons.


0

According to the diagram you provided, $|a\rangle$ is injected into a beam-splitter, resulting in the out-put state, \begin{equation} |\psi_{o}\rangle =B|a\rangle= e^{\frac{i\pi}{2}}|b\rangle +|c\rangle. \end{equation} Here the beam-splitter unitary is denoted by $B$. I have expressed $i$ as $e^{\frac{i\pi}{2}}$, to highlight the fact that $|b\rangle$ picks ...


0

In the 2 glasses there are 4 surfaces, i.e interface air/glass, and 8 surface orientations (a..h) and plenty room for interference between reflections and the main beam. LASER (air) a1b (glass) c2d (air) e3f (glass) g4h At each interface the is reflection that will be reflected forward again (self-interference) look for iridiscence in ...


1

At a guess, the effect rises from the fact that your interferometer is not properly aligned. The presence of linear, rather than circular, fringes suggests that there is an angular misalignment. Then moving the wedge causes a lateral shift in the intersection point of the beam and the angled slide, which results in a shift in the apparent position of the ...


1

This answer is based on Floris' insight that the slides might be bent. Let's say the laser hits the slide at an angle of $\theta$ and travels through the panel at an angle of $\theta'=\sin^{-1}({n_a\over n_s}\sin(\theta))$. Let's assume the curvature is light enough that the laser essentially exits parallel to how it entered. I am also going to assume you ...


3

This is speculating - but if your slides are of non-uniform thickness, or they are bent as a result of the pinching, they will present a different path length in one leg of the interferometer (and therefore give rise to a shift in the fringe pattern). This may become clear by looking at this diagram: In the diagram on the left, the total path length is ...



Top 50 recent answers are included